Mixing
Find exact value of tan when given cos
$begingroup$
Given $cos30 = frac{sqrt3}{2}$ use trigonometric identities to find the exact value of $tanfrac{pi}{3}$
I understand that $cos30 = frac{sqrt3}{2}$ from the standard trig values chart and I know that $frac{pi}{3}$ is 60 degrees and I know the value of it from the same chart. I'm not understanding how to use identities to find the value.
trigonometry
asked Jan 19 at 22:03
dstarhdstarh
1426
$endgroup$
add a comment |
$begingroup$
Given $cos30 = frac{sqrt3}{2}$ use trigonometric identities to find the exact value of $tanfrac{pi}{3}$
I understand that $cos30 = frac{sqrt3}{2}$ from the standard trig values chart and I know that $frac{pi}{3}$ is 60 degrees and I know the value of it from the same chart. I'm not understanding how to use identities to find the value.
trigonometry
asked Jan 19 at 22:03
dstarhdstarh
1426
$endgroup$
4
$begingroup$
What is "The" fundamental trigonometric identity? Once you know the sine of the angle you can easily compute the tangent.
$endgroup$
– Chris Leary
Jan 19 at 22:06
2
$begingroup$
I don't understand where the "$tan$" part in the question title comes in based on what is written in the question text. Is something missing?
$endgroup$
– John Omielan
Jan 19 at 22:07
2
$begingroup$
Given your title, are you rather asking, in the body of your post, to find the exact value of $tanleft(pi/3right)$?
$endgroup$
– jordan_glen
Jan 19 at 22:07
$begingroup$
Sorry missed writing $tan$ of
$endgroup$
– dstarh
Jan 19 at 22:08
add a comment |
$begingroup$
Given $cos30 = frac{sqrt3}{2}$ use trigonometric identities to find the exact value of $tanfrac{pi}{3}$
I understand that $cos30 = frac{sqrt3}{2}$ from the standard trig values chart and I know that $frac{pi}{3}$ is 60 degrees and I know the value of it from the same chart. I'm not understanding how to use identities to find the value.
trigonometry
asked Jan 19 at 22:03
dstarhdstarh
1426
$endgroup$
Given $cos30 = frac{sqrt3}{2}$ use trigonometric identities to find the exact value of $tanfrac{pi}{3}$
I understand that $cos30 = frac{sqrt3}{2}$ from the standard trig values chart and I know that $frac{pi}{3}$ is 60 degrees and I know the value of it from the same chart. I'm not understanding how to use identities to find the value.
trigonometry
trigonometry
asked Jan 19 at 22:03
dstarhdstarh
1426
asked Jan 19 at 22:03
dstarhdstarh
1426
edited Jan 19 at 22:07
dstarh
asked Jan 19 at 22:03
dstarhdstarh
1426
asked Jan 19 at 22:03
dstarhdstarh
1426
asked Jan 19 at 22:03
dstarhdstarh
1426
1426
4
$begingroup$
What is "The" fundamental trigonometric identity? Once you know the sine of the angle you can easily compute the tangent.
$endgroup$
– Chris Leary
Jan 19 at 22:06
2
$begingroup$
I don't understand where the "$tan$" part in the question title comes in based on what is written in the question text. Is something missing?
$endgroup$
– John Omielan
Jan 19 at 22:07
2
$begingroup$
Given your title, are you rather asking, in the body of your post, to find the exact value of $tanleft(pi/3right)$?
$endgroup$
– jordan_glen
Jan 19 at 22:07
$begingroup$
Sorry missed writing $tan$ of
$endgroup$
– dstarh
Jan 19 at 22:08
add a comment |
4
$begingroup$
What is "The" fundamental trigonometric identity? Once you know the sine of the angle you can easily compute the tangent.
$endgroup$
– Chris Leary
Jan 19 at 22:06
2
$begingroup$
I don't understand where the "$tan$" part in the question title comes in based on what is written in the question text. Is something missing?
$endgroup$
– John Omielan
Jan 19 at 22:07
2
$begingroup$
Given your title, are you rather asking, in the body of your post, to find the exact value of $tanleft(pi/3right)$?
$endgroup$
– jordan_glen
Jan 19 at 22:07
$begingroup$
Sorry missed writing $tan$ of
$endgroup$
– dstarh
Jan 19 at 22:08
4
4
$begingroup$
What is "The" fundamental trigonometric identity? Once you know the sine of the angle you can easily compute the tangent.
$endgroup$
– Chris Leary
Jan 19 at 22:06
$begingroup$
What is "The" fundamental trigonometric identity? Once you know the sine of the angle you can easily compute the tangent.
$endgroup$
– Chris Leary
Jan 19 at 22:06
2
2
$begingroup$
I don't understand where the "$tan$" part in the question title comes in based on what is written in the question text. Is something missing?
$endgroup$
– John Omielan
Jan 19 at 22:07
$begingroup$
I don't understand where the "$tan$" part in the question title comes in based on what is written in the question text. Is something missing?
$endgroup$
– John Omielan
Jan 19 at 22:07
2
2
$begingroup$
Given your title, are you rather asking, in the body of your post, to find the exact value of $tanleft(pi/3right)$?
$endgroup$
– jordan_glen
Jan 19 at 22:07
$begingroup$
Given your title, are you rather asking, in the body of your post, to find the exact value of $tanleft(pi/3right)$?
$endgroup$
– jordan_glen
Jan 19 at 22:07
$begingroup$
Sorry missed writing $tan$ of
$endgroup$
– dstarh
Jan 19 at 22:08
$begingroup$
Sorry missed writing $tan$ of
$endgroup$
– dstarh
Jan 19 at 22:08
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Upto the sign, you can calculate the tan-value as $$tan(x)=frac{sqrt{1-cos^2(x)}}{cos(x)}$$
Also note $$cos(2x)=2cos^2(x)-1$$ which allows you to calculate $cos(frac{pi}{3})$ from $cos(frac{pi}{6})$
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
$endgroup$
2
$begingroup$
But the question is asking $tanfrac{pi}{3}$ and you have $cos frac{pi}{6}$
$endgroup$
– user289143
Jan 19 at 22:12
2
$begingroup$
But $x$ is not $pi/6$, but rather $pi/3$.
$endgroup$
– jordan_glen
Jan 19 at 22:13
add a comment |
$begingroup$
Now $$tanfrac{pi}{3}=frac{sinfrac{pi}{3}}{cosfrac{pi}{3}}=frac{cosfrac{pi}{6}}{sin frac{pi}{6}}=frac{cos frac{pi}{6}}{sqrt{1-cos^2frac{pi}{6}}}=frac{frac{sqrt{3}}{2}}{sqrt{1-frac{3}{4}}}=frac{sqrt{3}}{2}cdot frac{1}{sqrt{frac{1}{4}}}=sqrt{3}$$
answered Jan 19 at 22:11
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
Just to use another elementary way, which shows all the passages:
$$cos(60) = cos(30+30) = cos(30)cos(30) - sin(30)sin(30)$$
The value of $cos(30)$ you know it.
The value of $sin(30)$ is derived from $cos^2 + sin^2 = 1$.
Again, once you found $cos(60)$ you get the value of $sin(60)$ for free.
Now you can find $tan(60)$.
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
$endgroup$
add a comment |
$begingroup$
begin{align}
&color{red}{tanleft(fracpi3right)}=frac{sinleft(fracpi3right)}{cosleft(fracpi3right)}=frac{sqrt{1-cos^2left(fracpi3right)}}{cosleft(fracpi3right)}=frac{sqrt{1-left(2color{blue}{cosleft(fracpi6right)}^2-1right)^2}}{2color{blue}{cosleft(fracpi6right)}^2-1}\
&=frac{sqrt{1-left(2left(color{blue}{frac{sqrt 3}2}right)^2-1right)^2}}{2left(color{blue}{frac{sqrt 3}2}right)^2-1}=frac{sqrt{1-left(2frac34-1right)^2}}{2frac34-1}=frac{sqrt{1-left(frac12right)^2}}{frac12}\
&=frac{sqrt{frac34}}{frac12}=frac{frac12sqrt3}{frac12}=color{red}{sqrt 3}
end{align}
$$therefore~tanleft(fracpi3right)=sqrt3$$
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
$endgroup$
add a comment |
$begingroup$
$sin(pi/3) = sqrt 3/2,$ and $cos(pi/3) = 1/2$.
Therefore, $$tan(pi/3) = dfrac{sinleft(frac pi 3right)}{cosleft(frac pi 3right)} = dfrac {dfrac{sqrt 3}2}{dfrac 12}=sqrt 3$$
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
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add a comment |
$begingroup$
Assume $x = fracpi6$, thus $fracpi3 = 2x$
$$tanleft(2xright) = frac{sinleft(2xright)}{cosleft(2xright)} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1}$$
Since, $cosleft(xright)=cosleft(fracpi6right)=frac{sqrt 3}2$, substituting that on above equation, you'd get:
$$tanleft(2xright) = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-frac 34right)}left(frac{sqrt{3}}2right)}{2times frac 34-1} = sqrt{3}$$
$$therefore~tanleft(fracpi3right)=sqrt3$$
Trigonometric Identities used are:
$$sinleft(2xright) = 2sinleft(xright)cosleft(xright)$$
$$cosleft(2xright) = 2cos^2left(xright)-1$$
$$cos^2left(xright)+sin^2left(xright)=1$$
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
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add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Upto the sign, you can calculate the tan-value as $$tan(x)=frac{sqrt{1-cos^2(x)}}{cos(x)}$$
Also note $$cos(2x)=2cos^2(x)-1$$ which allows you to calculate $cos(frac{pi}{3})$ from $cos(frac{pi}{6})$
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
$endgroup$
2
$begingroup$
But the question is asking $tanfrac{pi}{3}$ and you have $cos frac{pi}{6}$
$endgroup$
– user289143
Jan 19 at 22:12
2
$begingroup$
But $x$ is not $pi/6$, but rather $pi/3$.
$endgroup$
– jordan_glen
Jan 19 at 22:13
add a comment |
$begingroup$
Upto the sign, you can calculate the tan-value as $$tan(x)=frac{sqrt{1-cos^2(x)}}{cos(x)}$$
Also note $$cos(2x)=2cos^2(x)-1$$ which allows you to calculate $cos(frac{pi}{3})$ from $cos(frac{pi}{6})$
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
$endgroup$
2
$begingroup$
But the question is asking $tanfrac{pi}{3}$ and you have $cos frac{pi}{6}$
$endgroup$
– user289143
Jan 19 at 22:12
2
$begingroup$
But $x$ is not $pi/6$, but rather $pi/3$.
$endgroup$
– jordan_glen
Jan 19 at 22:13
add a comment |
$begingroup$
Upto the sign, you can calculate the tan-value as $$tan(x)=frac{sqrt{1-cos^2(x)}}{cos(x)}$$
Also note $$cos(2x)=2cos^2(x)-1$$ which allows you to calculate $cos(frac{pi}{3})$ from $cos(frac{pi}{6})$
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
$endgroup$
Upto the sign, you can calculate the tan-value as $$tan(x)=frac{sqrt{1-cos^2(x)}}{cos(x)}$$
Also note $$cos(2x)=2cos^2(x)-1$$ which allows you to calculate $cos(frac{pi}{3})$ from $cos(frac{pi}{6})$
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
edited Jan 19 at 22:24
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
answered Jan 19 at 22:08
PeterPeter
48.1k1139133
48.1k1139133
2
$begingroup$
But the question is asking $tanfrac{pi}{3}$ and you have $cos frac{pi}{6}$
$endgroup$
– user289143
Jan 19 at 22:12
2
$begingroup$
But $x$ is not $pi/6$, but rather $pi/3$.
$endgroup$
– jordan_glen
Jan 19 at 22:13
add a comment |
2
$begingroup$
But the question is asking $tanfrac{pi}{3}$ and you have $cos frac{pi}{6}$
$endgroup$
– user289143
Jan 19 at 22:12
2
$begingroup$
But $x$ is not $pi/6$, but rather $pi/3$.
$endgroup$
– jordan_glen
Jan 19 at 22:13
2
2
$begingroup$
But the question is asking $tanfrac{pi}{3}$ and you have $cos frac{pi}{6}$
$endgroup$
– user289143
Jan 19 at 22:12
$begingroup$
But the question is asking $tanfrac{pi}{3}$ and you have $cos frac{pi}{6}$
$endgroup$
– user289143
Jan 19 at 22:12
2
2
$begingroup$
But $x$ is not $pi/6$, but rather $pi/3$.
$endgroup$
– jordan_glen
Jan 19 at 22:13
$begingroup$
But $x$ is not $pi/6$, but rather $pi/3$.
$endgroup$
– jordan_glen
Jan 19 at 22:13
add a comment |
$begingroup$
Now $$tanfrac{pi}{3}=frac{sinfrac{pi}{3}}{cosfrac{pi}{3}}=frac{cosfrac{pi}{6}}{sin frac{pi}{6}}=frac{cos frac{pi}{6}}{sqrt{1-cos^2frac{pi}{6}}}=frac{frac{sqrt{3}}{2}}{sqrt{1-frac{3}{4}}}=frac{sqrt{3}}{2}cdot frac{1}{sqrt{frac{1}{4}}}=sqrt{3}$$
answered Jan 19 at 22:11
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
Now $$tanfrac{pi}{3}=frac{sinfrac{pi}{3}}{cosfrac{pi}{3}}=frac{cosfrac{pi}{6}}{sin frac{pi}{6}}=frac{cos frac{pi}{6}}{sqrt{1-cos^2frac{pi}{6}}}=frac{frac{sqrt{3}}{2}}{sqrt{1-frac{3}{4}}}=frac{sqrt{3}}{2}cdot frac{1}{sqrt{frac{1}{4}}}=sqrt{3}$$
answered Jan 19 at 22:11
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
Now $$tanfrac{pi}{3}=frac{sinfrac{pi}{3}}{cosfrac{pi}{3}}=frac{cosfrac{pi}{6}}{sin frac{pi}{6}}=frac{cos frac{pi}{6}}{sqrt{1-cos^2frac{pi}{6}}}=frac{frac{sqrt{3}}{2}}{sqrt{1-frac{3}{4}}}=frac{sqrt{3}}{2}cdot frac{1}{sqrt{frac{1}{4}}}=sqrt{3}$$
answered Jan 19 at 22:11
user289143user289143
1,002313
$endgroup$
Now $$tanfrac{pi}{3}=frac{sinfrac{pi}{3}}{cosfrac{pi}{3}}=frac{cosfrac{pi}{6}}{sin frac{pi}{6}}=frac{cos frac{pi}{6}}{sqrt{1-cos^2frac{pi}{6}}}=frac{frac{sqrt{3}}{2}}{sqrt{1-frac{3}{4}}}=frac{sqrt{3}}{2}cdot frac{1}{sqrt{frac{1}{4}}}=sqrt{3}$$
answered Jan 19 at 22:11
user289143user289143
1,002313
answered Jan 19 at 22:11
user289143user289143
1,002313
answered Jan 19 at 22:11
user289143user289143
1,002313
answered Jan 19 at 22:11
user289143user289143
1,002313
1,002313
add a comment |
add a comment |
$begingroup$
Just to use another elementary way, which shows all the passages:
$$cos(60) = cos(30+30) = cos(30)cos(30) - sin(30)sin(30)$$
The value of $cos(30)$ you know it.
The value of $sin(30)$ is derived from $cos^2 + sin^2 = 1$.
Again, once you found $cos(60)$ you get the value of $sin(60)$ for free.
Now you can find $tan(60)$.
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
$endgroup$
add a comment |
$begingroup$
Just to use another elementary way, which shows all the passages:
$$cos(60) = cos(30+30) = cos(30)cos(30) - sin(30)sin(30)$$
The value of $cos(30)$ you know it.
The value of $sin(30)$ is derived from $cos^2 + sin^2 = 1$.
Again, once you found $cos(60)$ you get the value of $sin(60)$ for free.
Now you can find $tan(60)$.
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
$endgroup$
add a comment |
$begingroup$
Just to use another elementary way, which shows all the passages:
$$cos(60) = cos(30+30) = cos(30)cos(30) - sin(30)sin(30)$$
The value of $cos(30)$ you know it.
The value of $sin(30)$ is derived from $cos^2 + sin^2 = 1$.
Again, once you found $cos(60)$ you get the value of $sin(60)$ for free.
Now you can find $tan(60)$.
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
$endgroup$
Just to use another elementary way, which shows all the passages:
$$cos(60) = cos(30+30) = cos(30)cos(30) - sin(30)sin(30)$$
The value of $cos(30)$ you know it.
The value of $sin(30)$ is derived from $cos^2 + sin^2 = 1$.
Again, once you found $cos(60)$ you get the value of $sin(60)$ for free.
Now you can find $tan(60)$.
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
answered Jan 19 at 22:17
Von NeumannVon Neumann
16.4k72545
16.4k72545
add a comment |
add a comment |
$begingroup$
begin{align}
&color{red}{tanleft(fracpi3right)}=frac{sinleft(fracpi3right)}{cosleft(fracpi3right)}=frac{sqrt{1-cos^2left(fracpi3right)}}{cosleft(fracpi3right)}=frac{sqrt{1-left(2color{blue}{cosleft(fracpi6right)}^2-1right)^2}}{2color{blue}{cosleft(fracpi6right)}^2-1}\
&=frac{sqrt{1-left(2left(color{blue}{frac{sqrt 3}2}right)^2-1right)^2}}{2left(color{blue}{frac{sqrt 3}2}right)^2-1}=frac{sqrt{1-left(2frac34-1right)^2}}{2frac34-1}=frac{sqrt{1-left(frac12right)^2}}{frac12}\
&=frac{sqrt{frac34}}{frac12}=frac{frac12sqrt3}{frac12}=color{red}{sqrt 3}
end{align}
$$therefore~tanleft(fracpi3right)=sqrt3$$
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
$endgroup$
add a comment |
$begingroup$
begin{align}
&color{red}{tanleft(fracpi3right)}=frac{sinleft(fracpi3right)}{cosleft(fracpi3right)}=frac{sqrt{1-cos^2left(fracpi3right)}}{cosleft(fracpi3right)}=frac{sqrt{1-left(2color{blue}{cosleft(fracpi6right)}^2-1right)^2}}{2color{blue}{cosleft(fracpi6right)}^2-1}\
&=frac{sqrt{1-left(2left(color{blue}{frac{sqrt 3}2}right)^2-1right)^2}}{2left(color{blue}{frac{sqrt 3}2}right)^2-1}=frac{sqrt{1-left(2frac34-1right)^2}}{2frac34-1}=frac{sqrt{1-left(frac12right)^2}}{frac12}\
&=frac{sqrt{frac34}}{frac12}=frac{frac12sqrt3}{frac12}=color{red}{sqrt 3}
end{align}
$$therefore~tanleft(fracpi3right)=sqrt3$$
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
$endgroup$
add a comment |
$begingroup$
begin{align}
&color{red}{tanleft(fracpi3right)}=frac{sinleft(fracpi3right)}{cosleft(fracpi3right)}=frac{sqrt{1-cos^2left(fracpi3right)}}{cosleft(fracpi3right)}=frac{sqrt{1-left(2color{blue}{cosleft(fracpi6right)}^2-1right)^2}}{2color{blue}{cosleft(fracpi6right)}^2-1}\
&=frac{sqrt{1-left(2left(color{blue}{frac{sqrt 3}2}right)^2-1right)^2}}{2left(color{blue}{frac{sqrt 3}2}right)^2-1}=frac{sqrt{1-left(2frac34-1right)^2}}{2frac34-1}=frac{sqrt{1-left(frac12right)^2}}{frac12}\
&=frac{sqrt{frac34}}{frac12}=frac{frac12sqrt3}{frac12}=color{red}{sqrt 3}
end{align}
$$therefore~tanleft(fracpi3right)=sqrt3$$
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
$endgroup$
begin{align}
&color{red}{tanleft(fracpi3right)}=frac{sinleft(fracpi3right)}{cosleft(fracpi3right)}=frac{sqrt{1-cos^2left(fracpi3right)}}{cosleft(fracpi3right)}=frac{sqrt{1-left(2color{blue}{cosleft(fracpi6right)}^2-1right)^2}}{2color{blue}{cosleft(fracpi6right)}^2-1}\
&=frac{sqrt{1-left(2left(color{blue}{frac{sqrt 3}2}right)^2-1right)^2}}{2left(color{blue}{frac{sqrt 3}2}right)^2-1}=frac{sqrt{1-left(2frac34-1right)^2}}{2frac34-1}=frac{sqrt{1-left(frac12right)^2}}{frac12}\
&=frac{sqrt{frac34}}{frac12}=frac{frac12sqrt3}{frac12}=color{red}{sqrt 3}
end{align}
$$therefore~tanleft(fracpi3right)=sqrt3$$
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
answered Jan 19 at 22:24
mrtaurhomrtaurho
5,59051440
5,59051440
add a comment |
add a comment |
$begingroup$
$sin(pi/3) = sqrt 3/2,$ and $cos(pi/3) = 1/2$.
Therefore, $$tan(pi/3) = dfrac{sinleft(frac pi 3right)}{cosleft(frac pi 3right)} = dfrac {dfrac{sqrt 3}2}{dfrac 12}=sqrt 3$$
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
$sin(pi/3) = sqrt 3/2,$ and $cos(pi/3) = 1/2$.
Therefore, $$tan(pi/3) = dfrac{sinleft(frac pi 3right)}{cosleft(frac pi 3right)} = dfrac {dfrac{sqrt 3}2}{dfrac 12}=sqrt 3$$
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
$sin(pi/3) = sqrt 3/2,$ and $cos(pi/3) = 1/2$.
Therefore, $$tan(pi/3) = dfrac{sinleft(frac pi 3right)}{cosleft(frac pi 3right)} = dfrac {dfrac{sqrt 3}2}{dfrac 12}=sqrt 3$$
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
$endgroup$
$sin(pi/3) = sqrt 3/2,$ and $cos(pi/3) = 1/2$.
Therefore, $$tan(pi/3) = dfrac{sinleft(frac pi 3right)}{cosleft(frac pi 3right)} = dfrac {dfrac{sqrt 3}2}{dfrac 12}=sqrt 3$$
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
edited Jan 19 at 22:16
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
answered Jan 19 at 22:11
jordan_glenjordan_glen
1
1
add a comment |
add a comment |
$begingroup$
Assume $x = fracpi6$, thus $fracpi3 = 2x$
$$tanleft(2xright) = frac{sinleft(2xright)}{cosleft(2xright)} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1}$$
Since, $cosleft(xright)=cosleft(fracpi6right)=frac{sqrt 3}2$, substituting that on above equation, you'd get:
$$tanleft(2xright) = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-frac 34right)}left(frac{sqrt{3}}2right)}{2times frac 34-1} = sqrt{3}$$
$$therefore~tanleft(fracpi3right)=sqrt3$$
Trigonometric Identities used are:
$$sinleft(2xright) = 2sinleft(xright)cosleft(xright)$$
$$cosleft(2xright) = 2cos^2left(xright)-1$$
$$cos^2left(xright)+sin^2left(xright)=1$$
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
$endgroup$
add a comment |
$begingroup$
Assume $x = fracpi6$, thus $fracpi3 = 2x$
$$tanleft(2xright) = frac{sinleft(2xright)}{cosleft(2xright)} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1}$$
Since, $cosleft(xright)=cosleft(fracpi6right)=frac{sqrt 3}2$, substituting that on above equation, you'd get:
$$tanleft(2xright) = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-frac 34right)}left(frac{sqrt{3}}2right)}{2times frac 34-1} = sqrt{3}$$
$$therefore~tanleft(fracpi3right)=sqrt3$$
Trigonometric Identities used are:
$$sinleft(2xright) = 2sinleft(xright)cosleft(xright)$$
$$cosleft(2xright) = 2cos^2left(xright)-1$$
$$cos^2left(xright)+sin^2left(xright)=1$$
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
$endgroup$
add a comment |
$begingroup$
Assume $x = fracpi6$, thus $fracpi3 = 2x$
$$tanleft(2xright) = frac{sinleft(2xright)}{cosleft(2xright)} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1}$$
Since, $cosleft(xright)=cosleft(fracpi6right)=frac{sqrt 3}2$, substituting that on above equation, you'd get:
$$tanleft(2xright) = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-frac 34right)}left(frac{sqrt{3}}2right)}{2times frac 34-1} = sqrt{3}$$
$$therefore~tanleft(fracpi3right)=sqrt3$$
Trigonometric Identities used are:
$$sinleft(2xright) = 2sinleft(xright)cosleft(xright)$$
$$cosleft(2xright) = 2cos^2left(xright)-1$$
$$cos^2left(xright)+sin^2left(xright)=1$$
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
$endgroup$
Assume $x = fracpi6$, thus $fracpi3 = 2x$
$$tanleft(2xright) = frac{sinleft(2xright)}{cosleft(2xright)} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sinleft(xright)cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1}$$
Since, $cosleft(xright)=cosleft(fracpi6right)=frac{sqrt 3}2$, substituting that on above equation, you'd get:
$$tanleft(2xright) = frac{2sqrt{left(1-cos^2left(xright)right)}cosleft(xright)}{2cos^2left(xright)-1} = frac{2sqrt{left(1-frac 34right)}left(frac{sqrt{3}}2right)}{2times frac 34-1} = sqrt{3}$$
$$therefore~tanleft(fracpi3right)=sqrt3$$
Trigonometric Identities used are:
$$sinleft(2xright) = 2sinleft(xright)cosleft(xright)$$
$$cosleft(2xright) = 2cos^2left(xright)-1$$
$$cos^2left(xright)+sin^2left(xright)=1$$
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
edited Jan 19 at 23:56
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
answered Jan 19 at 23:46
Mathew MahindaratneMathew Mahindaratne
536212
536212
add a comment |
add a comment |
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4
$begingroup$
What is "The" fundamental trigonometric identity? Once you know the sine of the angle you can easily compute the tangent.
$endgroup$
– Chris Leary
Jan 19 at 22:06
2
$begingroup$
I don't understand where the "$tan$" part in the question title comes in based on what is written in the question text. Is something missing?
$endgroup$
– John Omielan
Jan 19 at 22:07
2
$begingroup$
Given your title, are you rather asking, in the body of your post, to find the exact value of $tanleft(pi/3right)$?
$endgroup$
– jordan_glen
Jan 19 at 22:07
$begingroup$
Sorry missed writing $tan$ of
$endgroup$
– dstarh
Jan 19 at 22:08